The unit's digit in the product `(7^(71)xx6^(63)xx3^(65))` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the unit's digit in the product \(7^{71} \times 6^{63} \times 3^{65}\), we can follow these steps:

### Step 1: Determine the unit's digit of each term in the product.
1. **For \(7^{71}\)**:
- The unit's digits of powers of 7 follow a cycle:
- \(7^1 = 7\) (unit digit is 7)
- \(7^2 = 49\) (unit digit is 9)
- \(7^3 = 343\) (unit digit is 3)
- \(7^4 = 2401\) (unit digit is 1)
- The cycle is \(7, 9, 3, 1\) and repeats every 4 terms.
- To find the position in the cycle for \(7^{71}\), calculate \(71 \mod 4\):
\[ 71 \div 4 = 17 \quad \text{(remainder 3)} \]
- So, \(71 \mod 4 = 3\). The unit's digit of \(7^{71}\) is the same as that of \(7^3\), which is **3**.

2. **For \(6^{63}\)**:
- The unit's digit of any power of 6 is always **6** (since \(6^1 = 6\), \(6^2 = 36\), etc.).
- Thus, the unit's digit of \(6^{63}\) is **6**.

3. **For \(3^{65}\)**:
- The unit's digits of powers of 3 also follow a cycle:
- \(3^1 = 3\) (unit digit is 3)
- \(3^2 = 9\) (unit digit is 9)
- \(3^3 = 27\) (unit digit is 7)
- \(3^4 = 81\) (unit digit is 1)
- The cycle is \(3, 9, 7, 1\) and repeats every 4 terms.
- To find the position in the cycle for \(3^{65}\), calculate \(65 \mod 4\):
\[ 65 \div 4 = 16 \quad \text{(remainder 1)} \]
- So, \(65 \mod 4 = 1\). The unit's digit of \(3^{65}\) is the same as that of \(3^1\), which is **3**.

### Step 2: Combine the unit's digits.
Now we have:
- Unit's digit of \(7^{71}\) is **3**.
- Unit's digit of \(6^{63}\) is **6**.
- Unit's digit of \(3^{65}\) is **3**.

We need to find the unit's digit of the product:
\[ 3 \times 6 \times 3 \]

### Step 3: Calculate the product step by step.
1. First, calculate \(3 \times 6\):
\[ 3 \times 6 = 18 \quad \text{(unit digit is 8)} \]
2. Next, multiply the result by 3:
\[ 8 \times 3 = 24 \quad \text{(unit digit is 4)} \]

### Final Answer:
The unit's digit in the product \(7^{71} \times 6^{63} \times 3^{65}\) is **4**.

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